Titration calculations and their uncertainty

The standard route

1. Moles of the solution with known concentration: n = c x V (V in dm³).

2. Use the balanced equation to find moles of the other substance.

3. Scale up if a sample (aliquot) was taken: 25.0 cm³ from 250.0 cm³ means x 10.

4. Find what is asked: concentration (n / V), molar mass (mass / n), percentage purity (mass found / mass weighed x 100) or water of crystallisation.

c1V1 = c2V2 works only for a 1 : 1 reaction ratio. Common errors are using the wrong ratio and forgetting to scale up (October 2024 report).

Water of crystallisation and purity

• x in Na2CO3·xH2O: find M of the hydrated salt from mass / n, subtract M of the anhydrous salt and divide by 18.0. Round to the nearest whole number.
• Purity: moles from the titration x M gives the mass of pure substance in the sample; divide by the mass weighed.

Uncertainty in the result

Add the percentage uncertainties of the balance, volumetric flask, pipette and titre. The titre usually contributes most when it is small, so larger titres give a more reliable result.

Worked example

2.65 g of anhydrous Na2CO3 is made up to 250.0 cm³. 25.0 cm³ portions need a mean titre of 24.30 cm³ of hydrochloric acid. Find [HCl].

1. n(Na2CO3) total = 2.65 / 106.0 = 0.02500 mol; in 25.0 cm³ = 0.002500 mol
2. Na2CO3 + 2HCl -> 2NaCl + H2O + CO2, so n(HCl) = 2 x that
3. [HCl] = 0.005000 / 0.02430

Answer: 0.206 mol dm^-3

Check yourself

1. 25.0 cm³ of sodium hydroxide solution needs 22.50 cm³ of 0.100 mol dm^-3 sulfuric acid. Calculate the concentration of the sodium hydroxide.
   Show answer

   0.180 mol dm^-3. n(H2SO4) = 0.02250 x 0.100 = 2.25 x 10^-3 mol; n(NaOH) = 2 x that = 4.50 x 10^-3 mol; c = 4.50 x 10^-3 / 0.0250 = 0.180 mol dm^-3.

2. 1.50 g of a solid monoprotic acid is made up to 250.0 cm³. 25.0 cm³ portions need 20.40 cm³ of 0.0500 mol dm^-3 NaOH. Calculate the molar mass of the acid.
   Show answer

   147 g mol^-1. n(NaOH) = 0.02040 x 0.0500 = 1.02 x 10^-3 mol = n(acid) in 25.0 cm³; total = 1.02 x 10^-2 mol; M = 1.50 / 0.0102 = 147 g mol^-1.

3. 3.58 g of Na2CO3·xH2O is made up to 250.0 cm³. 25.0 cm³ portions need 25.00 cm³ of 0.100 mol dm^-3 HCl. Find x (Na2CO3 = 106.0, H2O = 18.0).
   Show answer

   10. n(HCl) = 2.50 x 10^-3 mol; n(Na2CO3) = 1.25 x 10^-3 mol per portion, 0.0125 mol in total; M = 3.58 / 0.0125 = 286.4; x = (286.4 - 106.0) / 18.0 = 10.02, so x = 10.

4. 2.50 g of impure sulfamic acid (monoprotic, M = 97.1 g mol^-1) is made up to 250.0 cm³. 25.0 cm³ portions need 24.80 cm³ of 0.100 mol dm^-3 NaOH. Calculate the percentage purity.
   Show answer

   96.3 %. n per portion = 2.48 x 10^-3 mol; total 0.0248 mol; mass = 0.0248 x 97.1 = 2.408 g; purity = 96.3%.

5. A titre is 20.40 cm³ (two readings, each ±0.05 cm³). Calculate its percentage uncertainty.
   Show answer

   0.49 %. 0.10 / 20.40 x 100 = 0.49%.

6. When is it correct to use c1V1 = c2V2 to find an unknown concentration from a titration?
   • For every acid-base titration
   • Only when the reacting ratio is 1 : 1
   • Only for sulfuric acid
   • Only when the titre is 25.00 cm³
   Show answer

   Only when the reacting ratio is 1 : 1. Otherwise the ratio from the equation must be included.

7. A student forgets to multiply by 10 after titrating 25.0 cm³ portions from a 250.0 cm³ flask, when finding the molar mass of a solid. What happens to their answer?
   • It is 10 times too small
   • It is unchanged
   • It becomes negative
   • It is 10 times too large
   Show answer

   It is 10 times too large. The moles used are 10 times too small, and M = mass / n, so M comes out 10 times too big.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson IAL Chemistry teacher practical guide
• Pearson examiners' report, WCH16 October 2024
• Pearson examiners' report, WCH13 October 2021

Last updated 8 November 2026. Spotted a mistake? Tell us.